 Hello friends, this is Dheeraj welcoming you again on another problem solving video. So this question was asked by one of our students. So let us discuss about this question and try to learn some light. So let us first read the question. So you can see there's a satellite of mass m which is launched vertically upward with velocity u from the surface of the earth. After it reaches the height of r, the height is measured from the surface. Always remember that. Now this height is also the radius of the earth. After reaching it ejects a rocket of mass m by 10. And immediately after it, the satellite starts moving in a circular orbit. So we need to find out the kinetic energy of the rocket. So you know, if you do a lot of problem solving yourself, you realize that this particular question will probably take up your at least four to five minutes. So in case you're writing a J main exam and this kind of question comes up in during your first, let us say first 15 or 20 minutes. So I would advise you to mark this question to solve it later. Otherwise this question can really kill all your time. All right. So but right now we are here to learn the concept. So let us proceed. So let us say this is the earth. OK, now there's a rocket of mass m that is launched vertically from here. OK, so let us say it is launched like that with a velocity of u. And once it reaches a radius r, so I'll take a lot of time to explain. All right, so that we can get it properly. So after let us say it reaches a height r from the surface of the earth, where r is a radius of the earth also. After reaching here, this mass m splits into two parts. What happens? It ejects a rocket of mass m by 10. OK, so remaining mass of the satellite becomes 9 m by 10. And then satellite starts moving in a circular orbit. So right now when it is here, when it is about to eject the rocket, it was going radially outwards. Isn't it? And immediately after, let us say when it ejects a rocket, let us say it ejects a rocket like this. I'll put a point here. Let us say this small red dot is a rocket. This rocket has been ejected. Now you can see the direction of ejection. It cannot be ejected in, let's say horizontal or vertical direction. The reason is the satellite has to gain a tangential velocity. Let us say the satellite's velocity just after the ejection of the rocket is V. Now, can it have a radial component, my dear friends? Of course not. If there is a radial component, the satellite will keep going away, right? But it has to immediately start revolving in a circle. So that's why immediately after the ejection of the rocket, the satellite's radial component of the velocity or you can say the component of velocity in the y direction where this is the y direction and this is the x direction. So immediately after the ejection of the rocket, the satellite's y direction momentum becomes zero, momentum or velocity. Now this gives us a good amount of hint here. So what you have to do is, since there is a splitting of mass happening here and the rocket is getting ejected, you can definitely conserve momentum along both x as well as y direction. All right? So let us do one thing, the rocket which is getting ejected, we will assume its velocity along the x-axis and along the y-axis. We are basically taking the components of the velocity. So the rocket's velocity along the y-axis, let us say it is Vy and the rocket's velocity along the x-axis is let us say Vx. All right? So keep noticing all these terms that we are using. V is the velocity of the satellite after the rocket has ejected and the mass of the satellite has become 9m by 10 because m by 10 is the mass of the rocket itself. All right? So let us proceed from here till the point the rocket is getting ejected. I can conserve the energy. Right? So I can say u1 plus k1 is equal to u2 plus k2. This is the conservation of mechanical energy equation, fine? So I am going to take mass of the earth as m, which they want you to take. So that is why they are given it. And g is the gravitational constant, r is also given, my dear friends, all right? So in this equation, we will just have to substitute u1 is what? Minus of gm into small m by r, this is point number 1, all right? And this is point number 2. Point number 2 is just before the ejection of the rocket, fine? This plus kinetic energy, half m into u2. That is equal to potential g at point number 2, which is minus of gm into m divided by 2r. So rocket just about to come out, plus k2, half m into, let us say, velocity is v1 for the rocket when, sorry, for the satellite when the rocket just about to come out of the velocity of the satellite is v1, all right? So you are on the right track, don't worry. So this is the velocity in which direction along the y direction for the satellite just before the rocket is about to eject out of it. Now what we will do? We will conserve the momentum along the x-axis and along the y-axis. So conservation of linear momentum, my dear friends, we have to use along the x-axis. How will I use it? I will write along the x-axis, there was no momentum initially for the satellite, isn't it? That is equal to m by 10 into velocity of the satellite vx minus, sorry, velocity of the rocket vx minus 9m by 10, which is the remaining satellite's mass, that into v, fine? So this is the conservation of linear momentum along the x-axis. Now the equation along the y-axis, along the y-axis, you will see that immediately after the ejection of the rocket, the remaining portion of the satellite should not have any velocity on the y-axis because it has to move in a circle. So entire velocity have to be tangential. I hope you are getting my point here. So along the y-axis, initial momentum is m into v1, which is nothing but m into root over u square minus gm by r. This is your momentum of the satellite along with the rocket before rocket came out. That should be equal to the final momentum, right? So the rocket's final momentum along the y-axis is m by 10 vy and the remaining portion of the satellite does not have any momentum along the y-axis as immediately after it starts revolving in a circle. So these are the two equations, fine? So from here, you can, from the first equation, you will get the value of vx, vx is equal to 9 times v. Now you can basically arrange the term in any which way. It entirely depends upon what the question is asking, right? The question is asking kinetic energy of the rocket and what is the kinetic energy of the rocket? The kinetic energy of the rocket is nothing but half mass of the rocket, which is m by 10 into vx square plus vy square, right? So this bracket term is a total velocity square of the rocket. So that is what we have to find out. So we have found out vx. Now we'll be finding vy. So vy is equal to 10 times root over u square minus gm by r, right? So you have vx and vy. But the problem is, the problem is that the vx is given in terms of an unknown quantity v. So do we know v? So if you think carefully, you will understand that the fact that the satellite is revolving at a distance of 2r, its velocity is automatically fixed. It has to have certain tangential velocity. So let us find out how we can proceed from here on. So whenever something is moving in a server orbit, we tend to write force towards the center is equal to mv square by r, right? So I hope all of that you remember. So gm into mass of the rocket, which is m by 10 divided by the distance square, which is 2r square, that should be equal to m by 10 into v square divided by radius. So this is nothing but force towards the center is equal to mv square by r. You can probably use some direct expressions also. But don't you worry if you know the basics. I think I have written small m in terms of capital M. So I will replace it. This is capital M. So small m by 10 goes away. One of the 2r is also gone. So from here, you will get v is equal to root over gm by 2r. And you can get vx is equal to 9 times v, which is 9 times root over gm by 2r. So my dear friends, here is your second equation. Now you have now vy and vx, the kinetic energy of the rocket, which is asked is simply equal to half m by 10 into, let me write over here, into vx square plus vy square. Now although we have solved the question just for the clarity of the concepts. But if such kind of question, let's say you get in J main, you need to immediately identify that this question could eat up a lot of your time. So you need to skip it as soon as possible. And I'm not saying that you should skip it and don't come back at the end in case you have time left or whatever simple or easy questions you might have been getting. If you have solved all of them, then you can definitely come back to this question and solve. All right. So thanks for watching the video. I hope you have learned something today. Bye bye.